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Distance between two parallel planes - class-XII
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Distance between the parallel planes $2x-3y+4z-1=0$ and $4x-6y+8z+8=0$ is
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A
$\dfrac{5}{\sqrt{29}}$
💡 Explanation:
Consider the given line
Let,
$2x-3y+4z-1=0$ ----- $(1)$
And,
$4x - 6y + 8z + 8 = 0$
$2x - 3y + 4z + 4 = 0$ ---- $(2)$
Now,
Distance between plane $1$ and $2$
$d=|\dfrac{d _1-d _2}{\sqrt {a^2+b^2+c^2}}|$
$=|\dfrac{-1-4}{\sqrt {2^2+(-3)^2+4^2}}|=\dfrac{5}{\sqrt {4+9+16}}$
$=\dfrac{5}{\sqrt {29}}$
Hence, distance between the planes is $\dfrac{5}{\sqrt {29}}$
So,
Option $A$ is correct.